A quadrilateral has angle measures of 84°, 107°, and 58°.

1 answer:

3 0

Hello!

The angles in a quadrilateral add to 360°

Subtract the known angles from 360 to get the missing angle

360 - 84 - 107 - 58 = 111

The answer is 111°

Hope this helps!

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One day you bike for 30 miles 10 cool down at a rate of 0.25 miles/minute. The total distance you bike is 35 miles. How many min

tatyana61 [14]

To determine the length of time it would took for the biker to cool down, we need the rate that would relate the distance he traveled to cool down per units of time. For this problem, the rate is given as 0.25 miles per minute. So, we simply divide the total distance he traveled with this rate. We calculate as follows:

time to cool down = distance / rate
time to cool down = 35 miles / 0.25 miles / minute
time to cool down = 140 minutes or 2 hrs and 20 minutes

Therefore, the biker would need to cool down for about 2 hrs and 20 minutes if he traveled for 35 miles.

8 0

3 years ago

Ben and Josh went to the roof of their 40-foot tall high school to throw their math books offthe edge.The initial velocity of Be

Taya2010 [7]

Answer

Josh's textbook reached the ground first

Josh's textbook reached the ground first by a difference of $t=0.6482$

Step-by-step explanation:

Before we proceed let us re write correctly the height equation which in correct form reads:

$h(t)=-16t^2 +v_{o}t+s$ Eqn(1).

Where:

$h(t)$ : is the height range as a function of time

$v_{o}$ : is the initial velocity

$s$ : is the initial heightin feet and is given as 40 feet, thus Eqn(1). becomes:

$h(t)=-16t^2 + v_{o}t + 40$ Eqn(2).

Now let us use the given information and set up our equations for Ben and Josh.

Ben:

We know that $v_{o}=60ft/s$

Thus Eqn. (2) becomes:

$h(t)=-16t^2+60t+40$ Eqn.(3)

Josh:

We know that $v_{o}=48ft/s$

Thus Eqn. (2) becomes:

$h(t)=-16t^2+48t+40$ Eqn. (4).

Now since we want to find whose textbook reaches the ground first and by how many seconds we need to solve each equation (i.e. Eqns. (3) and (4)) at $h(t)=0$ . Now since both are quadratic equations we will solve one showing the full method which can be repeated for the other one.

Thus we have for Ben, Eqn. (3) gives:

$h(t)=0-16t^2+60t+40=0$

Using the quadratic expression to find the roots of the quadratic we have:

$t_{1,2}=\frac{-b+/-\sqrt{b^2-4ac} }{2a} \\t_{1,2}=\frac{-60+/-\sqrt{60^2-4(-16)(40)} }{2(-16)} \\t_{1,2}=\frac{-60+/-\sqrt{6160} }{-32} \\t_{1,2}=\frac{15+/-\sqrt{385} }{8}\\\\t_{1}=4.3276 sec\\t_{2}=-0.5776 sec$